Question

What will be the intermediate sight distance on a highway for a design speed of 100 kmph when its descending gradient is 3%? Assuming coefficient of friction as 0.6 and brake efficiency as 60%.

Answer 1

$SSD=Vt=\frac{V^2}{2g(f\pm n)}$

$V=0.278\times 100m/s=27.8m/s$

$S=9.81m/s^2,f=0.6\times 0.6=0.36$
$\because \text{brake efficiency}=60\%$

$n=0.03$

For descending gradient
$SSD=vt+\frac{v^2}{2g(f-n)}$
$=27.8\times 2.5+\frac{{27.8}^2}{2\times 9.81(0.36-0.03)}$
$=188.86m$

$ISD=2SSD$
So,
$ISD=377.7m$